Find m such that the equation has only 1 solution. | Mathematics | Page 1

fleet craterBOT Jul 7, 2024, 12:31 PM

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novel apex Jul 7, 2024, 3:06 PM

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delta =0 is not a sufficient condition

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say in a case delta>0 for some 'm'

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it will not necessarily have 2 roots

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because the expected root can be negative and 2^x is not negative
so think of satisfactory condition

thin quartz Jul 7, 2024, 10:17 PM

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could you write the equation in latex

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the one you wanna solve

flat basin Jul 7, 2024, 10:21 PM

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$(m-2)4^{x}+(2m-3)2^{1+x}+5m-6 = 0$

late leafBOT Jul 7, 2024, 10:21 PM

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Ravi (No Lifer)

flat basin Jul 7, 2024, 10:21 PM

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Only one value for x

waxen topaz Jul 7, 2024, 11:22 PM

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its this

novel apex Jul 8, 2024, 8:47 AM

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hmm
see, if there has to be one root
one root of quadratic should be negative and one should be positive
so, product of roots<0

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i hope that makes sense

flat basin Jul 8, 2024, 3:34 PM

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late leaf **Ravi (No Lifer)**

wait

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lmfao

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$2^x = a$

late leafBOT Jul 8, 2024, 3:34 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:34 PM

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$(m-2)a^2 + 2(2m-3)a + 5m-6 = 0$

late leafBOT Jul 8, 2024, 3:34 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:50 PM

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so b^2 - 4ac = 0

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$4(2m-3)^2 - 4(m-2)(5m-6) = 0$

late leafBOT Jul 8, 2024, 3:50 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:51 PM

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$4m^2 + 9 - 12m - (5m^2 - 6m - 10m + 12) =0$

late leafBOT Jul 8, 2024, 3:51 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:51 PM

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$-m^2 + 4m -3 = 0$

late leafBOT Jul 8, 2024, 3:52 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:52 PM

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$m^2 - 3m - m + 3 = 0$

late leafBOT Jul 8, 2024, 3:52 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:52 PM

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$(m-1)(m-3) = 0$

late leafBOT Jul 8, 2024, 3:52 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:55 PM

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wtf

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ok inequality bash ig

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$(m-2)4^{x}+(2m-3)2^{1+x}+5m-6 = 0$

late leafBOT Jul 8, 2024, 3:57 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:58 PM

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$m4^x - 2^{2x+1} + m2^{2+x} - 3\times2^{1+x} + 5m - 6 =0$

late leafBOT Jul 8, 2024, 3:58 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 3:59 PM

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$m2^{2x} + m2^{2+x} + 5m = 2^{2x+1} + 3\times2^{1+x} + 6$

late leafBOT Jul 8, 2024, 3:59 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:00 PM

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$m(2^{2x} + 2^{2+x} + 5) = 2(2^{2x} + 3\times2^x+3)$

late leafBOT Jul 8, 2024, 4:00 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:02 PM

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huh what if

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$2^x = \frac{2(3-2m)\pm\sqrt{4(2m-3)^2-4(m-2)(5m-6)}}{2(m-2)}$

late leafBOT Jul 8, 2024, 4:04 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:04 PM

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$2^x = 3-2m\pm\sqrt{(2m-3)^2-(m-2)(5m-6)}$

late leafBOT Jul 8, 2024, 4:04 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:05 PM

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$0 \leq 3-2m\pm\sqrt{(2m-3)^2-(m-2)(5m-6)}$

late leafBOT Jul 8, 2024, 4:05 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:06 PM

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$2m-3\leq\pm\sqrt{(2m-3)^2-(m-2)(5m-6)}$

late leafBOT Jul 8, 2024, 4:06 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:07 PM

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$0<-(m-2)(5m-6)$

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$0< -(5m^2-6m-10m+12)$

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$0 < -5m^2 + 16m - 12$

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Let $\alpha,\beta$ be the roots of $-5m^2+16m-12$

Assume $\alpha \leq \beta$

the solution for $0\leq -5m^2+16m-12$ is $m\in(\alpha,\beta)$

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,w -5m^2+16m-12=0

late leafBOT Jul 8, 2024, 4:11 PM

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Ravi (No Lifer)

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Ravi (No Lifer)

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Ravi (No Lifer)

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:12 PM

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$\alpha = \frac65$

$\beta = 2$

late leafBOT Jul 8, 2024, 4:12 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:13 PM

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$m\in(1.2,2)$

late leafBOT Jul 8, 2024, 4:13 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:13 PM

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and we are done.

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$$2^x = \frac{2(3-2m)\pm\sqrt{4(2m-3)^2-4(m-2)(5m-6)}}{2(m-2)}$$

$$2^x = 3-2m\pm\sqrt{(2m-3)^2-(m-2)(5m-6)}$$

$$0 < 3-2m\pm\sqrt{(2m-3)^2-(m-2)(5m-6)}$$

$$2m-3<\pm\sqrt{(2m-3)^2-(m-2)(5m-6)}$$

$$0<-(m-2)(5m-6)$$

$$0< -(5m^2-6m-10m+12)$$

$$0 < -5m^2 + 16m - 12$$

Let $\alpha,\beta$ be the roots of $-5m^2+16m-12$

Assume $\alpha \leq \beta$

the solution for $0\leq -5m^2+16m-12$ is $m\in(\alpha,\beta)$

$$m\in(1.2,2)$$

late leafBOT Jul 8, 2024, 4:16 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:16 PM

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late leaf **Ravi (No Lifer)**

@thin quartz here you go

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nope

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at m=1.2 the zero occurs at -infty

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and at m=2 probably +infty

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lmfao graph it

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check for 1.2 and 2

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there isn't a zero in R

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maybe in $\overline{\bR}$

late leafBOT Jul 8, 2024, 4:21 PM

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Ravi (No Lifer)

flat basin Jul 8, 2024, 4:21 PM

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but we don't care for thar

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*that

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lmao instead of bashing conditions

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use the fact that for a real value of x to exist, 2^x > 0

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that is the easiet method tbh

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real x

half estuaryBOT Jul 8, 2024, 4:37 PM

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@little marlin has given 1 rep to @novel apex @flat basin

flat basin Jul 8, 2024, 4:51 PM

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Sure lmao

thin quartz Jul 8, 2024, 10:08 PM

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for what

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raaah

flat basin Jul 9, 2024, 1:19 AM

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thin quartz raaah

We shall shitpost here

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novel apex Jul 9, 2024, 3:00 AM

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flat basin $$2^x = \frac{2(3-2m)\pm\sqrt{4(2m-3)^2-4(m-2)(5m-6)}}{2(m-2)}$$ $$2^x = 3-2m\p...

umm you are not supposed to solve the whole problem while helping

thin quartz Jul 9, 2024, 10:39 AM

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+close

#Find m such that the equation has only 1 solution.